Problem: There is a light source on the plane. This source is so small that it can be represented as point. The light source is moving from point (a, s_y) to the (b, s_y) (s_y < 0) with speed equal to 1 unit per second. The trajectory of this light source is a straight segment connecting these two points. 

There is also a fence on OX axis represented as n segments (l_i, r_i) (so the actual coordinates of endpoints of each segment are (l_i, 0) and (r_i, 0)). The point (x, y) is in the shade if segment connecting (x,y) and the current position of the light source intersects or touches with any segment of the fence.

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You are given q points. For each point calculate total time of this point being in the shade, while the light source is moving from (a, s_y) to the (b, s_y).

Input

First line contains three space separated integers s_y, a and b (-10^9 ≤ s_y < 0, 1 ≤ a < b ≤ 10^9) — corresponding coordinates of the light source.

Second line contains single integer n (1 ≤ n ≤ 2 ⋅ 10^5) — number of segments in the fence.

Next n lines contain two integers per line: l_i and r_i (1 ≤ l_i < r_i ≤ 10^9, r_{i - 1} < l_i) — segments in the fence in increasing order. Segments don't intersect or touch each other.

Next line contains single integer q (1 ≤ q ≤ 2 ⋅ 10^5) — number of points to check.

Next q lines contain two integers per line: x_i and y_i (1 ≤ x_i, y_i ≤ 10^9) — points to process.

Output

Print q lines. The i-th line should contain one real number — total time of the i-th point being in the shade, while the light source is moving from (a, s_y) to the (b, s_y). The answer is considered as correct if its absolute of relative error doesn't exceed 10^{-6}.

Example

Input

-3 1 6
2
2 4
6 7
5
3 1
1 3
6 1
6 4
7 6


Output

5.000000000000000
3.000000000000000
0.000000000000000
1.500000000000000
2.000000000000000

Note

  * The 1-st point is always in the shade; 
  * the 2-nd point is in the shade while light source is moving from (3, -3) to (6, -3); 
  * the 3-rd point is in the shade while light source is at point (6, -3). 
  * the 4-th point is in the shade while light source is moving from (1, -3) to (2.5, -3) and at point (6, -3); 
  * the 5-th point is in the shade while light source is moving from (1, -3) to (2.5, -3) and from (5.5, -3) to (6, -3); 

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
long double e = 1e-9;
int main() {
  ios::sync_with_stdio(0), cin.tie(0), cout.tie(0);
  long long s, a, b;
  cin >> s >> a >> b;
  int n;
  cin >> n;
  long long l[n], r[n];
  for (int i = 0; i < n; i++) {
    cin >> l[i] >> r[i];
  }
  auto doca = [&](long double xx, long double x, long double y) {
    return xx - xx * (s / y) + x * (s / y) - a;
  };
  long long pref[n];
  pref[0] = r[0] - l[0];
  for (int i = 1; i < n; i++) {
    pref[i] = pref[i - 1] + r[i] - l[i];
  }
  int q;
  cin >> q;
  for (int i = 0; i < q; i++) {
    long long x, y;
    cin >> x >> y;
    int low = 0, high = n - 1;
    if (doca(r[high], x, y) <= 0 || doca(l[low], x, y) >= (b - a)) {
      cout << 0 << "\n";
      continue;
    }
    while (low < high) {
      int mid = (low + high) / 2;
      if (doca(r[mid], x, y) > 0) {
        high = mid;
      } else {
        low = mid + 1;
      }
    }
    long double ans = 0;
    if (doca(l[high], x, y) < 0) {
      if (doca(r[high], x, y) > (b - a)) {
        cout << b - a << "\n";
        continue;
      }
      ans += doca(r[high], x, y);
      high++;
    }
    long long lt = high;
    if (!(lt >= n)) {
      long double cons = y - s;
      cons /= y;
      low = high;
      high = n - 1;
      if (doca(l[low], x, y) > (b - a)) {
        goto prnt;
      }
      while (low < high) {
        int mid = (low + high + 1) / 2;
        if (doca(l[mid], x, y) <= (b - a)) {
          low = mid;
        } else {
          high = mid - 1;
        }
      }
      long long rt = low;
      if (doca(r[high], x, y) > (b - a)) {
        ans += (long double)(b - a) - doca(l[high], x, y);
        rt--;
      }
      if (lt <= rt) {
        ans += (pref[rt] - ((lt == 0) ? 0 : pref[lt - 1])) * cons;
      }
    }
  prnt:
    cout << fixed << setprecision(12) << ans << "\n";
  }
  return 0;
}